# 1988 USAMO Problems/Problem 2

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## Problem

The cubic polynomial $x^3+ax^2+bx+c$ has real coefficients and three real roots $r\ge s\ge t$. Show that $k=a^2-3b\ge 0$ and that $\sqrt k\le r-t$.

## Solution

### Solution 1

By Vieta's Formulas, $a=-r-s-t$, $b=rs+st+rt$, and $c=-rst$. Now we know $k=a^2-3b$; in terms of r, s, and t, then, $$k=(-r-s-t)^2-3(rs+st+rt)$$ $$k=r^2+s^2+t^2-rs-st-rt$$ Now notice that we can multiply both sides by 2, and rearrange terms to get $2k=(r-s)^2+(s-t)^2+(r-t)^2$. But since $r, s, t\in \mathbb{R}$, the three terms of the RHS are all non-negative (as the square of a real number is always non-negative), and therefore their sum is also non-negative -- that is, $2k\ge 0 \Rightarrow k\ge 0$.

Now, we will show that $\sqrt k\le r-t$. We can square both sides, and the inequality will hold since they are both non-negative (it is given that $r\ge t$, therefore $r-t\ge 0$). This gives $k \le r^2-2rt+t^2$. Now we already have $k=r^2+s^2+t^2-rs-st-rt$, so substituting this for k gives $$r^2+s^2+t^2-rs-st-rt \le r^2-2rt+t^2$$ $$s^2-rs-st+rt \le 0$$ $$s^2-(r+t)s+rt \le 0$$ Note that this is a quadratic. Since its leading coefficient is positive, its value is less than 0 when s is between the two roots. Using the quadratic formula: $$s=\frac {r+t\pm \sqrt {(-r-t)^2-4rt} } 2$$ $$s=\frac {r+t\pm \sqrt {r^2-2rt+t^2} } 2$$ $$s=\frac {r+t\pm (r-t) } 2$$ $$s \in \{r, t\}$$ The quadratic is 0 when s is equal to r or t, and the inequality holds when its value is less than or equal to 0 -- that is, $r\ge s\ge t$. (Its value is less than or equal to 0 when s is between the roots, since the graph of the quadratic opens upward.) In fact, the problem tells us this is true. Q.E.D.

### Solution 2

From Vieta's Formula (which tells us that $a = -(r+s+t)$ and $b = rs + st + rt$), we have that $$k = a^2 - 3b = r^2 + s^2 + t^2 - rs - st - rt = \frac{1}{2} ((r-s)^2 + (s-t)^2 + (r-t)^2),$$ clearly non-negative. To prove $\sqrt{k} \le r - t$, it suffices to prove the square of this relation, or $$r^2 + s^2 + t^2 - rs - st - rt \le r^2 - 2rt + t^2.$$ This in turn simplifies to $$rs + st - rt - s^2 \ge 0,$$ or $$(r - s)(s - t) \ge 0,$$ which is clearly true as $r \ge s \ge t$. This completes the proof.

## Solution 3

By Vieta's Formulas, $a = -(r+s+t)$ and $b = rs + st + rt$. $k = (r+s+t)^2 - 3(rs + st + rt) = r^2+s^2+t^2-rs-st-rt = \frac{(r-s)^2 + (s-t)^2 + (r-t)^2}{2}$.

To show that $k \ge 0$, simply note that by the trivial inequality, all three squares are greater than $0$ as they are the squares of real numbers.

To show that $\sqrt{k} \le r-t$, since both are positive, it is sufficient to show that $k \le (r-t)^2$. $\frac{(r-s)^2 + (s-t)^2 + (r-t)^2}{2} \le (r-t)^2$ implies that $k \le (r-t)^2$. $\frac{(r-s)^2 + (s-t)^2 - (r-t)^2}{2} \le 0$. Let $y = r-s$ and $z = s-t$. We then have $\frac{y^2 + z^2 - (y+z)^2}{2} \le 0 \implies -2yz \le 0$, which is clearly true as both $y$ and $z$ are positive.

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